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"use strict";(self.webpackChunkfi=self.webpackChunkfi||[]).push([[7728],{73212:(s,e,a)=>{a.r(e),a.d(e,{assets:()=>c,contentTitle:()=>t,default:()=>d,frontMatter:()=>i,metadata:()=>m,toc:()=>r});var n=a(85893),l=a(11151);const i={id:"top-down-dp",slug:"/recursion/pyramid-slide-down/top-down-dp",title:"Top-down DP solution",description:"Top-down DP solution of the Pyramid Slide Down.\n",tags:["java","dynamic-programming","top-down-dp"],last_update:{date:new Date("2023-08-17T00:00:00.000Z")}},t="Top-down dynamic programming",m={id:"recursion/2023-08-17-pyramid-slide-down/top-down-dp",title:"Top-down DP solution",description:"Top-down DP solution of the Pyramid Slide Down.\n",source:"@site/algorithms/04-recursion/2023-08-17-pyramid-slide-down/03-top-down-dp.md",sourceDirName:"04-recursion/2023-08-17-pyramid-slide-down",slug:"/recursion/pyramid-slide-down/top-down-dp",permalink:"/algorithms/recursion/pyramid-slide-down/top-down-dp",draft:!1,unlisted:!1,editUrl:"https://github.com/mfocko/blog/tree/main/algorithms/04-recursion/2023-08-17-pyramid-slide-down/03-top-down-dp.md",tags:[{label:"java",permalink:"/algorithms/tags/java"},{label:"dynamic-programming",permalink:"/algorithms/tags/dynamic-programming"},{label:"top-down-dp",permalink:"/algorithms/tags/top-down-dp"}],version:"current",lastUpdatedAt:1692230400,formattedLastUpdatedAt:"Aug 17, 2023",sidebarPosition:3,frontMatter:{id:"top-down-dp",slug:"/recursion/pyramid-slide-down/top-down-dp",title:"Top-down DP solution",description:"Top-down DP solution of the Pyramid Slide Down.\n",tags:["java","dynamic-programming","top-down-dp"],last_update:{date:"2023-08-17T00:00:00.000Z"}},sidebar:"autogeneratedBar",previous:{title:"Greedy solution",permalink:"/algorithms/recursion/pyramid-slide-down/greedy"},next:{title:"Bottom-up DP solution",permalink:"/algorithms/recursion/pyramid-slide-down/bottom-up-dp"}},c={},r=[{value:"Time complexity",id:"time-complexity",level:2},{value:"Memory complexity",id:"memory-complexity",level:2}];function h(s){const e={a:"a",admonition:"admonition",annotation:"annotation",code:"code",em:"em",h1:"h1",h2:"h2",li:"li",math:"math",mi:"mi",mn:"mn",mo:"mo",mrow:"mrow",mspace:"mspace",mstyle:"mstyle",msub:"msub",mtable:"mtable",mtd:"mtd",mtr:"mtr",munderover:"munderover",ol:"ol",p:"p",pre:"pre",section:"section",semantics:"semantics",span:"span",strong:"strong",sup:"sup",...(0,l.a)(),...s.components};return(0,n.jsxs)(n.Fragment,{children:[(0,n.jsx)(e.h1,{id:"top-down-dynamic-programming",children:"Top-down dynamic programming"}),"\n",(0,n.jsxs)(e.p,{children:[(0,n.jsx)(e.em,{children:"Top-down dynamic programming"})," is probably the most common approach, since (at\nleast looks like) is the easiest to implement. The whole point is avoiding the\nunnecessary computations that we have already done."]}),"\n",(0,n.jsxs)(e.p,{children:["In our case, we can use our na\xefve solution and put a ",(0,n.jsx)(e.em,{children:"cache"})," on top of it that\nwill make sure, we don't do unnecessary calculations."]}),"\n",(0,n.jsx)(e.pre,{children:(0,n.jsx)(e.code,{className:"language-java",children:"// This \u201cstructure\u201d is required, since I have decided to use \u2039TreeMap\u203a which\n// requires the ordering on the keys. It represents one position in the pyramid.\nrecord Position(int row, int col) implements Comparable<Position> {\n public int compareTo(Position r) {\n if (row != r.row) {\n return Integer.valueOf(row).compareTo(r.row);\n }\n\n if (col != r.col) {\n return Integer.valueOf(col).compareTo(r.col);\n }\n\n return 0;\n }\n}\n\npublic static int longestSlideDown(\n int[][] pyramid,\n TreeMap<Position, Integer> cache,\n Position position) {\n int row = position.row;\n int col = position.col;\n\n if (row >= pyramid.length || col < 0 || col >= pyramid[row].length) {\n // BASE: out of bounds\n return Integer.MIN_VALUE;\n }\n\n if (row == pyramid.length - 1) {\n // BASE: bottom of the pyramid\n return pyramid[position.row][position.col];\n }\n\n if (!cache.containsKey(position)) {\n // We haven't computed the position yet, so we run the same \u201cformula\u201d as\n // in the na\xefve version \xbband\xab we put calculated slide into the cache.\n // Next time we want the slide down from given position, it will be just\n // retrieved from the cache.\n int slideDown = Math.max(\n longestSlideDown(pyramid, cache, new Position(row + 1, col)),\n longestSlideDown(pyramid, cache, new Position(row + 1, col + 1)));\n cache.put(position, pyramid[row][col] + slideDown);\n }\n\n return cache.get(position);\n}\n\npublic static int longestSlideDown(int[][] pyramid) {\n // At the beginning we need to create a cache and share it across the calls.\n TreeMap<Position, Integer> cache = new TreeMap<>();\n return longestSlideDown(pyramid, cache, new Position(0, 0));\n}\n"})}),"\n",(0,n.jsxs)(e.p,{children:["You have probably noticed that ",(0,n.jsx)(e.code,{children:"record Position"})," have appeared. Since we are\ncaching the already computed values, we need a \u201creasonable\u201d key. In this case we\nshare the cache only for one ",(0,n.jsx)(e.em,{children:"run"})," (i.e. pyramid) of the ",(0,n.jsx)(e.code,{children:"longestSlideDown"}),", so\nwe can cache just with the indices within the pyramid, i.e. the ",(0,n.jsx)(e.code,{children:"Position"}),"."]}),"\n",(0,n.jsx)(e.admonition,{title:"Record",type:"tip",children:(0,n.jsxs)(e.p,{children:[(0,n.jsx)(e.em,{children:"Record"})," is relatively new addition to the Java language. It is basically an\nimmutable structure with implicitly defined ",(0,n.jsx)(e.code,{children:".equals()"}),", ",(0,n.jsx)(e.code,{children:".hashCode()"}),",\n",(0,n.jsx)(e.code,{children:".toString()"})," and getters for the attributes."]})}),"\n",(0,n.jsxs)(e.p,{children:["Because of the choice of ",(0,n.jsx)(e.code,{children:"TreeMap"}),", we had to additionally define the ordering\non it."]}),"\n",(0,n.jsxs)(e.p,{children:["In the ",(0,n.jsx)(e.code,{children:"longestSlideDown"})," you can notice that the computation which used to be\nat the end of the na\xefve version above, is now wrapped in an ",(0,n.jsx)(e.code,{children:"if"})," statement that\nchecks for the presence of the position in the cache and computes the slide down\njust when it's needed."]}),"\n",(0,n.jsx)(e.h2,{id:"time-complexity",children:"Time complexity"}),"\n",(0,n.jsx)(e.p,{children:"If you think that evaluating time complexity for this approach is a bit more\ntricky, you are right. Keeping the cache in mind, it is not the easiest thing\nto do. However there are some observations that might help us figure this out:"}),"\n",(0,n.jsxs)(e.ol,{children:["\n",(0,n.jsx)(e.li,{children:"Slide down from each position is calculated only once."}),"\n",(0,n.jsx)(e.li,{children:"Once calculated, we use the result from the cache."}),"\n"]}),"\n",(0,n.jsxs)(e.p,{children:["Knowing this, we still cannot, at least easily, describe the time complexity of\nfinding the best slide down from a specific position, ",(0,n.jsx)(e.strong,{children:"but"})," we can bound it\nfrom above for the ",(0,n.jsx)(e.strong,{children:"whole"})," run from the top. Now the question is how we can do\nthat!"]}),"\n",(0,n.jsxs)(e.p,{children:["Overall we are doing the same things for almost",(0,n.jsx)(e.sup,{children:(0,n.jsx)(e.a,{href:"#user-content-fn-1",id:"user-content-fnref-1","data-footnote-ref":!0,"aria-describedby":"footnote-label",children:"1"})})," all of the positions within\nthe pyramid:"]}),"\n",(0,n.jsxs)(e.ol,{children:["\n",(0,n.jsxs)(e.li,{children:["\n",(0,n.jsx)(e.p,{children:"We calculate and store it (using the partial results stored in cache). This\nis done only once."}),"\n",(0,n.jsxs)(e.p,{children:["For each calculation we take 2 values from the cache and insert one value.\nBecause we have chosen ",(0,n.jsx)(e.code,{children:"TreeMap"}),", these 3 operations have logarithmic time\ncomplexity and therefore this step is equivalent to ",(0,n.jsxs)(e.span,{className:"katex",children:[(0,n.jsx)(e.span,{className:"katex-mathml",children:(0,n.jsx)(e.math,{xmlns:"http://www.w3.org/1998/Math/MathML",children:(0,n.jsxs)(e.semantics,{children:[(0,n.jsxs)(e.mrow,{children:[(0,n.jsx)(e.mn,{children:"3"}),(0,n.jsx)(e.mo,{children:"\u22c5"}),(0,n.jsxs)(e.msub,{children:[(0,n.jsxs)(e.mrow,{children:[(0,n.jsx)(e.mi,{children:"log"}),(0,n.jsx)(e.mo,{children:"\u2061"})]}),(0,n.jsx)(e.mn,{children:"2"})]}),(0,n.jsx)(e.mi,{children:"n"})]}),(0,n.jsx)(e.annotation,{encoding:"application/x-tex",children:"3 \\cdot \\log_2{n}"})]})})}),(0,n.jsxs)(e.span,{className:"katex-html","aria-hidden":"true",children:[(0,n.jsxs)(e.span,{className:"base",children:[(0,n.jsx)(e.span,{className:"strut",style:{height:"0.6444em"}}),(0,n.jsx)(e.span,{className:"mord",children:"3"}),(0,n.jsx)(e.span,{className:"mspace",style:{marginRight:"0.2222em"}}),(0,n.jsx)(e.span,{className:"mbin",children:"\u22c5"}),(0,n.jsx)(e.span,{className:"mspace",style:{marginRight:"0.2222em"}})]}),(0,n.jsxs)(e.span,{className:"base",children:[(0,n.jsx)(e.span,{className:"strut",style:{height:"0.9386em",verticalAlign:"-0.2441em"}}),(0,n.jsxs)(e.span,{className:"mop",children:[(0,n.jsxs)(e.span,{className:"mop",children:["lo",(0,n.jsx)(e.span,{style:{marginRight:"0.01389em"},children:"g"})]}),(0,n.jsx)(e.span,{className:"msupsub",children:(0,n.jsxs)(e.span,{className:"vlist-t 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will be interleaved with the next step, therefore it is easier to keep the\nretrievals in the following point."}),"\n",(0,n.jsx)(e.admonition,{type:"caution",children:(0,n.jsx)(e.p,{children:"You might have noticed it's still not that easy, cause we're not having full\ncache right from the beginning, but the sum of those logarithms cannot be\nexpressed in a nice way, so taking the upper bound, i.e. expecting the cache\nto be full at all times, is the best option for nice and readable complexity\nof the whole approach."})}),"\n",(0,n.jsxs)(e.p,{children:["Our final upper bound of this work is therefore ",(0,n.jsxs)(e.span,{className:"katex",children:[(0,n.jsx)(e.span,{className:"katex-mathml",children:(0,n.jsx)(e.math,{xmlns:"http://www.w3.org/1998/Math/MathML",children:(0,n.jsxs)(e.semantics,{children:[(0,n.jsxs)(e.mrow,{children:[(0,n.jsxs)(e.msub,{children:[(0,n.jsxs)(e.mrow,{children:[(0,n.jsx)(e.mi,{children:"log"}),(0,n.jsx)(e.mo,{children:"\u2061"})]}),(0,n.jsx)(e.mn,{children:"2"})]}),(0,n.jsx)(e.mi,{children:"n"})]}),(0,n.jsx)(e.annotation,{encoding:"application/x-tex",children:"\\log_2{n}"})]})})}),(0,n.jsx)(e.span,{className:"katex-html","aria-hidden":"true",children:(0,n.jsxs)(e.span,{className:"base",children:[(0,n.jsx)(e.span,{className:"strut",style:{height:"0.9386em",verticalAlign:"-0.2441em"}}),(0,n.jsxs)(e.span,{className:"mop",children:[(0,n.jsxs)(e.span,{className:"mop",children:["lo",(0,n.jsx)(e.span,{style:{marginRight:"0.01389em"},children:"g"})]}),(0,n.jsx)(e.span,{className:"msupsub",children:(0,n.jsxs)(e.span,{className:"vlist-t 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Same as in first point, but only twice, so we\nget ",(0,n.jsxs)(e.span,{className:"katex",children:[(0,n.jsx)(e.span,{className:"katex-mathml",children:(0,n.jsx)(e.math,{xmlns:"http://www.w3.org/1998/Math/MathML",children:(0,n.jsxs)(e.semantics,{children:[(0,n.jsxs)(e.mrow,{children:[(0,n.jsx)(e.mn,{children:"2"}),(0,n.jsx)(e.mo,{children:"\u22c5"}),(0,n.jsxs)(e.msub,{children:[(0,n.jsxs)(e.mrow,{children:[(0,n.jsx)(e.mi,{children:"log"}),(0,n.jsx)(e.mo,{children:"\u2061"})]}),(0,n.jsx)(e.mn,{children:"2"})]}),(0,n.jsx)(e.mi,{children:"n"})]}),(0,n.jsx)(e.annotation,{encoding:"application/x-tex",children:"2 \\cdot 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",(0,n.jsx)(e.code,{children:"if"})," condition."]})}),"\n"]}),"\n"]}),"\n",(0,n.jsx)(e.p,{children:"Okay, we have evaluated work done for each of the cells in the pyramid and now\nwe need to put it together."}),"\n",(0,n.jsx)(e.p,{children:"Let's split the time complexity of our solution into two operands:"}),"\n",(0,n.jsx)(e.span,{className:"katex-display",children:(0,n.jsxs)(e.span,{className:"katex",children:[(0,n.jsx)(e.span,{className:"katex-mathml",children:(0,n.jsx)(e.math,{xmlns:"http://www.w3.org/1998/Math/MathML",display:"block",children:(0,n.jsxs)(e.semantics,{children:[(0,n.jsxs)(e.mrow,{children:[(0,n.jsx)(e.mi,{mathvariant:"script",children:"O"}),(0,n.jsx)(e.mo,{stretchy:"false",children:"("}),(0,n.jsx)(e.mi,{children:"r"}),(0,n.jsx)(e.mo,{children:"+"}),(0,n.jsx)(e.mi,{children:"s"}),(0,n.jsx)(e.mo,{stretchy:"false",children:")"})]}),(0,n.jsx)(e.annotation,{encoding:"application/x-tex",children:"\\mathcal{O}(r + s)"})]})})}),(0,n.jsxs)(e.span,{className:"katex-html","aria-hidden":"true",children:[(0,n.jsxs)(e.span,{className:"base",children:[(0,n.jsx)(e.span,{className:"strut",style:{height:"1em",verticalAlign:"-0.25em"}}),(0,n.jsx)(e.span,{className:"mord mathcal",style:{marginRight:"0.02778em"},children:"O"}),(0,n.jsx)(e.span,{className:"mopen",children:"("}),(0,n.jsx)(e.span,{className:"mord mathnormal",style:{marginRight:"0.02778em"},children:"r"}),(0,n.jsx)(e.span,{className:"mspace",style:{marginRight:"0.2222em"}}),(0,n.jsx)(e.span,{className:"mbin",children:"+"}),(0,n.jsx)(e.span,{className:"mspace",style:{marginRight:"0.2222em"}})]}),(0,n.jsxs)(e.span,{className:"base",children:[(0,n.jsx)(e.span,{className:"strut",style:{height:"1em",verticalAlign:"-0.25em"}}),(0,n.jsx)(e.span,{className:"mord 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",(0,n.jsxs)(e.span,{className:"katex",children:[(0,n.jsx)(e.span,{className:"katex-mathml",children:(0,n.jsx)(e.math,{xmlns:"http://www.w3.org/1998/Math/MathML",children:(0,n.jsxs)(e.semantics,{children:[(0,n.jsx)(e.mrow,{children:(0,n.jsx)(e.mi,{children:"s"})}),(0,n.jsx)(e.annotation,{encoding:"application/x-tex",children:"s"})]})})}),(0,n.jsx)(e.span,{className:"katex-html","aria-hidden":"true",children:(0,n.jsxs)(e.span,{className:"base",children:[(0,n.jsx)(e.span,{className:"strut",style:{height:"0.4306em"}}),(0,n.jsx)(e.span,{className:"mord mathnormal",children:"s"})]})})]})," will represent\nthe additional retrievals on top of the calculation."]}),"\n",(0,n.jsxs)(e.p,{children:["We calculate the values only ",(0,n.jsx)(e.strong,{children:"once"}),", therefore we can safely agree 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This is not entirely true, since we have included the\n",(0,n.jsx)(e.code,{children:".containsKey()"})," and ",(0,n.jsx)(e.code,{children:".get()"})," from the ",(0,n.jsx)(e.code,{children:"return"})," statement in the second part."]}),(0,n.jsx)(e.p,{children:"If we were to represent this more precisely, we could've gone with:"}),(0,n.jsx)(e.span,{className:"katex-display",children:(0,n.jsxs)(e.span,{className:"katex",children:[(0,n.jsx)(e.span,{className:"katex-mathml",children:(0,n.jsx)(e.math,{xmlns:"http://www.w3.org/1998/Math/MathML",display:"block",children:(0,n.jsxs)(e.semantics,{children:[(0,n.jsxs)(e.mtable,{rowspacing:"0.25em",columnalign:"right 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&= 3 \\cdot n \\cdot \\log{n} \\\\\ns &= 2 \\cdot n \\cdot \\log{n}\n\\end{align*}"})]})})}),(0,n.jsx)(e.span,{className:"katex-html","aria-hidden":"true",children:(0,n.jsxs)(e.span,{className:"base",children:[(0,n.jsx)(e.span,{className:"strut",style:{height:"3em",verticalAlign:"-1.25em"}}),(0,n.jsx)(e.span,{className:"mord",children:(0,n.jsxs)(e.span,{className:"mtable",children:[(0,n.jsx)(e.span,{className:"col-align-r",children:(0,n.jsxs)(e.span,{className:"vlist-t vlist-t2",children:[(0,n.jsxs)(e.span,{className:"vlist-r",children:[(0,n.jsxs)(e.span,{className:"vlist",style:{height:"1.75em"},children:[(0,n.jsxs)(e.span,{style:{top:"-3.91em"},children:[(0,n.jsx)(e.span,{className:"pstrut",style:{height:"3em"}}),(0,n.jsx)(e.span,{className:"mord",children:(0,n.jsx)(e.span,{className:"mord 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However we are expressing the complexity in the\nBachmann-Landau notation, so we care about the ",(0,n.jsx)(e.strong,{children:"upper bound"}),", not the exact\nnumber."]})}),"\n",(0,n.jsxs)(e.admonition,{title:"Can this be optimized?",type:"tip",children:[(0,n.jsx)(e.p,{children:"Yes, it can! Try to think about a way, how can you minimize the memory\ncomplexity of this approach. I'll give you a hint:"}),(0,n.jsx)(e.span,{className:"katex-display",children:(0,n.jsxs)(e.span,{className:"katex",children:[(0,n.jsx)(e.span,{className:"katex-mathml",children:(0,n.jsx)(e.math,{xmlns:"http://www.w3.org/1998/Math/MathML",display:"block",children:(0,n.jsxs)(e.semantics,{children:[(0,n.jsxs)(e.mrow,{children:[(0,n.jsx)(e.mi,{mathvariant:"script",children:"O"}),(0,n.jsx)(e.mo,{stretchy:"false",children:"("}),(0,n.jsx)(e.mi,{children:"r"}),(0,n.jsx)(e.mi,{children:"o"}),(0,n.jsx)(e.mi,{children:"w"}),(0,n.jsx)(e.mi,{children:"s"}),(0,n.jsx)(e.mo,{stretchy:"false",children:")"})]}),(0,n.jsx)(e.annotation,{encoding:"application/x-tex",children:"\\mathcal{O}(rows)"})]})})}),(0,n.jsx)(e.span,{className:"katex-html","aria-hidden":"true",children:(0,n.jsxs)(e.span,{className:"base",children:[(0,n.jsx)(e.span,{className:"strut",style:{height:"1em",verticalAlign:"-0.25em"}}),(0,n.jsx)(e.span,{className:"mord mathcal",style:{marginRight:"0.02778em"},children:"O"}),(0,n.jsx)(e.span,{className:"mopen",children:"("}),(0,n.jsx)(e.span,{className:"mord mathnormal",children:"ro"}),(0,n.jsx)(e.span,{className:"mord mathnormal",style:{marginRight:"0.02691em"},children:"w"}),(0,n.jsx)(e.span,{className:"mord mathnormal",children:"s"}),(0,n.jsx)(e.span,{className:"mclose",children:")"})]})})]})})]}),"\n",(0,n.jsxs)(e.section,{"data-footnotes":!0,className:"footnotes",children:[(0,n.jsx)(e.h2,{className:"sr-only",id:"footnote-label",children:"Footnotes"}),"\n",(0,n.jsxs)(e.ol,{children:["\n",(0,n.jsxs)(e.li,{id:"user-content-fn-1",children:["\n",(0,n.jsxs)(e.p,{children:["except the bottom row ",(0,n.jsx)(e.a,{href:"#user-content-fnref-1","data-footnote-backref":"","aria-label":"Back to reference 1",className:"data-footnote-backref",children:"\u21a9"})]}),"\n"]}),"\n"]}),"\n"]})]})}function d(s={}){const{wrapper:e}={...(0,l.a)(),...s.components};return e?(0,n.jsx)(e,{...s,children:(0,n.jsx)(h,{...s})}):h(s)}},11151:(s,e,a)=>{a.d(e,{Z:()=>m,a:()=>t});var n=a(67294);const l={},i=n.createContext(l);function t(s){const e=n.useContext(i);return n.useMemo((function(){return"function"==typeof s?s(e):{...e,...s}}),[e,s])}function m(s){let e;return e=s.disableParentContext?"function"==typeof s.components?s.components(l):s.components||l:t(s.components),n.createElement(i.Provider,{value:e},s.children)}}}]);
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